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(2x)(2x+2)=480
We move all terms to the left:
(2x)(2x+2)-(480)=0
We multiply parentheses
4x^2+4x-480=0
a = 4; b = 4; c = -480;
Δ = b2-4ac
Δ = 42-4·4·(-480)
Δ = 7696
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{7696}=\sqrt{16*481}=\sqrt{16}*\sqrt{481}=4\sqrt{481}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{481}}{2*4}=\frac{-4-4\sqrt{481}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{481}}{2*4}=\frac{-4+4\sqrt{481}}{8} $
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